# Why is the particular solution of $y'' - 4y' +3y = e^t$ not in the form of $Ae^t$

The particular solution $Y_p(t)$ of this problem is actually in the form of $Ae^tt$, but isn't it supposed to be $Ae^t$ ? Since there is no homogenous root = 0, why do we need to multiply $t$.

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You need to multiply by $t$ because $e^t$ is already a solution of the homogeneous equation.

$$r^2-4 r+3 = 0 \implies (r-1) (r-3) = 0 \implies y^{(H)} = A e^t + B e^{3 t}$$

Substituting $e^t$ as the particular solution will produce zero, which will be no help. The next logical thing to do, then, is to use $t \, e^t$.

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and if $te^t$ is already in the homogeneous solution, we will need to multiply by one more $t$ making it $t^2e^t$? – 40Plot Apr 3 '13 at 2:07
If both $e^t$ and $t e^t$, yes. The point is to avoid getting zero when you plug in the proposed inhomogeneous solution. – Ron Gordon Apr 3 '13 at 2:08
"both $e^t$ and $te^t$" meaning the homogenous equation looks like $Ay'' - By' + Cy = e^t + te^t$, then the particular solution will look like $t^2e^t$, am I correct? – 40Plot Apr 3 '13 at 2:12
The case I illustrated is the solution to $$y''-2 y'+y = e^t$$ In that case, yes, the particular solution would take the form $(1/2) t^2 e^t$. – Ron Gordon Apr 3 '13 at 2:15

The homogeneous solution is dictated by:

$m^2 - 4m +3 = 0$.

That gives one root that equals $1$ and gives one solution $= e^t$.

Clear?

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Yes indeed, it's clear ;-) – amWhy Apr 9 '13 at 0:13